French
DeutschFile:Mrs Miniver's Problem.svg
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Original file (SVG file, nominally 900 × 348 pixels, file size: 793 bytes)
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Summary
| Description | English: Three instances of Mrs. Miniver's problem, of arranging two circles so that their area of overlap equals the area of their symmetric difference. In each case the inner yellow area equals the total area of the surrounding blue regions. The left case shows two circles of equal areas, the right case shows one circle with twice the area of the other, and the middle case is intermediate between these two. |
| Date | |
| Source | Own work |
| Author | David Eppstein |
Licensing
I, the copyright holder of this work, hereby publish it under the following license:
  | This file is made available under the Creative Commons CC0 1.0 Universal Public Domain Dedication. |
| The person who associated a work with this deed has dedicated the work to the public domain by waiving all of their rights to the work worldwide under copyright law, including all related and neighboring rights, to the extent allowed by law. You can copy, modify, distribute and perform the work, even for commercial purposes, all without asking permission.
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Source code
from math import pi,asin,sin,cos,acos from PADS.SVG import * import sys # Circular segment formulas from https://en.wikipedia.org/wiki/Circular_segment def areaFromRadius(r): return pi*r**2 def twoCircleArea(r1,r2): return areaFromRadius(r1)+areaFromRadius(r2) def angleFromChord(r,c): return 2*asin(c/(2*r)) def areaFromAngle(r,theta): return r**2*(theta-sin(theta))/2 def areaFromChord(r,c): theta = angleFromChord(r,c) return areaFromAngle(r,theta) def distanceToChord(r,c): theta = angleFromChord(r,c) return r*cos(theta/2) def separationFromChord(r1,r2,c): return distanceToChord(r1,c)+distanceToChord(r2,c) # Bisection to solve Mrs. Miniver's problem within floating point accuracy def miniverSeparation(r1,r2): if r1 > r2: r1,r2 = r2,r1 # make sure small circle first totalArea = twoCircleArea(r1,r2) targetArea = totalArea/3 def miniverTest(x): # x = cos(chord on small circle), -1: outside, 1: inside chord = 2*r1*sin(acos(x)) area1 = areaFromChord(r1,chord) if x > 0: area1 = areaFromRadius(r1) - area1 area2 = areaFromChord(r2,chord) return area1 + area2 < targetArea lo,hi = -1.0,1.0 for i in range(100): mid = (hi+lo)/2 if miniverTest(mid): lo = mid else: hi = mid chord = 2*r1*sin(acos(mid)) return distanceToChord(r2,chord) - r1*mid # Make it into a nice picture # Monochromatic and with a loose frame for now; we'll fix it up later gridUnit = 150 targetArea = 100000.0 boundingBox = (6+4j)*gridUnit output = SVG(boundingBox,sys.stdout) output.group(fill="none",stroke="#000") for i in (0,1,2): ratio = 2.0**(i*0.25) areaWhenSmall = twoCircleArea(1.0,ratio) factor = (targetArea/areaWhenSmall)**0.5 r1,r2 = factor,factor*ratio s = miniverSeparation(r1,r2) o = (1+2j+2*i)*gridUnit output.circle(o + (s+r2-r1)*0.5j, r1) output.circle(o - (s+r1-r2)*0.5j, r2) output.ungroup() output.close() File history
Click on a date/time to view the file as it appeared at that time.
| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 19:35, 29 May 2022 | | 900 × 348 (793 bytes) | David Eppstein | Keep outer circles as circles for tighter svg coding; should be no visible change |
| 06:48, 29 May 2022 |  | 900 × 348 (1 KB) | David Eppstein | Uploaded own work with UploadWizard |
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